令人惊讶的是，没有人提到这是我的一次性脚本的快速版本:

>>> x = 'gfgfdAAA1234ZZZuijjk'
>>> x.split('AAA')[1].split('ZZZ')[0]
'1234'

使用PyParsing

import pyparsing as pp

word = pp.Word(pp.alphanums)

s = 'gfgfdAAA1234ZZZuijjk'
rule = pp.nestedExpr('AAA', 'ZZZ')
for match in rule.searchString(s):
    print(match)

收益率:

[[1234]]

text = 'I want to find a string between two substrings'
left = 'find a '
right = 'between two'

print(text[text.index(left)+len(left):text.index(right)])

给了

string

如果你想寻找多次出现的情况。

content ="Prefix_helloworld_Suffix_stuff_Prefix_42_Suffix_andsoon"
strings = []
for c in content.split('Prefix_'):
    spos = c.find('_Suffix')
    if spos!=-1:
        strings.append( c[:spos])
print( strings )

或者更快:

strings = [ c[:c.find('_Suffix')] for c in content.split('Prefix_') if c.find('_Suffix')!=-1 ]

你可以使用re模块:

>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
('1234,)

此外，您可以在波纹函数中找到所有的组合

s = 'Part 1. Part 2. Part 3 then more text'
def find_all_places(text,word):
    word_places = []
    i=0
    while True:
        word_place = text.find(word,i)
        i+=len(word)+word_place
        if i>=len(text):
            break
        if word_place<0:
            break
        word_places.append(word_place)
    return word_places
def find_all_combination(text,start,end):
    start_places = find_all_places(text,start)
    end_places = find_all_places(text,end)
    combination_list = []
    for start_place in start_places:
        for end_place in end_places:
            print(start_place)
            print(end_place)
            if start_place>=end_place:
                continue
            combination_list.append(text[start_place:end_place])
    return combination_list
find_all_combination(s,"Part","Part")

结果:

['Part 1. ', 'Part 1. Part 2. ', 'Part 2. ']